Concrete in Australia Vol 39 No 4 49
The following letters are in response to a technical note published in the last issue: Shear in simply-supported reinforced concrete
beams, by Colin Gurley, Concrete in Australia Volume 39, Issue 2, June 2013 (pp 17-23).
Further explanation for clarity
The author is to be commended for his efforts over many
years (decades) to encourage structural designers to think more
carefully about building robustness into their structures by
careful detailing and capacity design.
This technical note, presumably the first of a series, deals
with the shear design of a simply-supported beam carrying a
uniformly distributed load, using an “exact” yield-line solution
developed by the author. The method is straightforward and
logical, and lends itself to programming in Excel.
It might have been helpful to have an explanatory second
paragraph in section 2.0 to indicate where the beam came from.
This could be along the lines of: “The beam carries an ultimate
design load w* of 125 kN/m, producing a mid-span bending
moment M* of 439 kN.m . This moment is carried by 5/N24
bottom bars. To ensure that bending failure occurs in preference
to shear failure, shear stirrups will be designed to carry a load
wu equal to 156 kN/m – the load that produces a central
moment equal to Mu.”
In section 3.0, the anchorage length has been chosen as 1200
mm rather than the basic development length of 1111 mm in
accordance with AS3600-2009. When 1111 mm is used, the
spacing of stirrups increases to 114 mm for the first segment
rather than 102 mm calculated in section 7.0 . The writer
believes that a strength reduction factor of 0.8, rather than 0.7
for shear, would be more appropriate in this setting, which is
more akin to steel in tension of a strut-and-tie analysis (AS3600
Table 2.2 .4). This, along with 1111 mm as the development
length, would open up the stirrup spacing to 131 mm over a
length of 540 mm – a more palatable figure. The calculated
stirrup spacing in the second segment becomes 368 mm over
a length of 1195 mm (3/N10 legs) and in the third segment is
399 mm (2/N10 legs). Incidentally, AS3600-2009 cl 13.1 .2.4 is
a better reference to proportionality because it relates to bars in
tension – the relevant situation for this beam.
Use of a truss analogy with the end struts inclined at an angle
to match the tension capacity of the 300 mm anchorage of the
bottom bars produces similar stirrup spacing in the end regions:
136 mm with phi equal to 0.8 compared with 131 mm above.
The author hints at only a small extra cost to achieve more
robustness under abnormal events. Using the author’s method,
but with a development length of 1111 mm, phi equal to
0.8 and load w* of 125 kN/m, a stirrup spacing of 227 mm
is required in the end region compared with 131mm. For
example, with practical spacing of 220 mm and 130 mm over
a length of about 660 mm, this converts to six stirrups rather
than four, i.e. two additional stirrups at each end – not a high
price to pay for a large additional benefit.
Use of a more disciplined and consistent terminology would
assist the reader. All of the following appear: First hinge, First
hanger segment, First fragment & first hinge, Hinge 1, First
segment, First dogleg hinge; Second hanger segment = first
tooth, Second hinge and first tooth, Hinge 2; Third segment;
and Dogleg yield hinge.
W H Boyce, FIEAust
In over their heads
I have passed around your paper to fellow staff and we agree,
for the apprentice drafters enrolling in the structural or civil
engineering courses we believe that this content is going to be
too technical for the students who attend.
This sort of information is more suitable for university
students who would have a better understanding of and
appreciate the complexity of designing a beam.
It is the main reason why this should not be included in
the course and why engineering firms have apprentices attend
TAFE and not university. The last thing we all want is for our
apprentices to struggle through the courses.
Also, if I’m not mistaken, they have something similar in
the syllabus that is more suited to apprentices and technical
students.
Anthony Bayadi,
NSW CAD Manager, Aurecon
For loads at the top
of the beam
May I thank Colin Gurley for drawing my attention to his
treatment of shear in reinforced concrete (RC) beams. In the
following discussion I have reverted to consideration of loads
applied at the top of the beam, the most common if not the
most critical situation. I have also used an alternative notation
where S = the yield capacity of vertical shear reo per metre.
That is, if shear reo is provided such that
Asv/s=Vutanθ/dfy
S=Asvfy/s=Vutanθ/d
where θ is the inclination of the compression struts, and
is measured from the intersect point of the forces above the
reaction in Colin’s Figure 4 to the centre of the compression
block; not along the dotted line. This touches on the node
detailing at this point in the strut-tie model, which will also
define the point at which the anchorage must have taken place
(assumed in Colin’s example to be the face of the support).
I have also ignored strength reduction (φ) factors since the
intention is to demonstrate behaviour, not provide design rules.
Colin’s paper looks at the support condition initially, and
allows θ to be determined by the anchorage force developed at
the bearing face. This, in his example, has increased θ compared
with the value that would apply were a higher anchorage
force chosen, and provided for (e.g . by anchor plates as per cl.
DISCUSSION PAPER
49-56 - Discussion.indd 49
49-56 - Discussion.indd 49
25/11/13 4:15 PM
25/11/13 4:15 PM